What is the integral of x/(sqrt(x-4)) ?

The integral of x/(sqrt(x-4)) is 2(1/3 (x-4)^{3/2}+4sqrt(x-4))+C

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integral of x/(sqrt(x-4))

\int \frac{x}{x - 4} d x

2 (\frac{1}{3} (x - 4)^{\frac{3}{2}} + 4 x - 4) + C

Solution steps

\int \frac{x}{x - 4} d x

Apply u-substitution

: \int 2 (u^{2} + 4) d u

= \int 2 (u^{2} + 4) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int u^{2} + 4 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int u^{2} d u + \int 4 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 4 d u = 4 u

= 2 (\frac{u ^{3}}{3} + 4 u)

Substitute back

u = x - 4

= 2 (\frac{( x - 4 ) ^{3}}{3} + 4 x - 4)

Simplify

2 (\frac{( x - 4 ) ^{3}}{3} + 4 x - 4) : 2 (\frac{1}{3} (x - 4)^{\frac{3}{2}} + 4 x - 4)

= 2 (\frac{1}{3} (x - 4)^{\frac{3}{2}} + 4 x - 4)

Add a constant to the solution

= 2 (\frac{1}{3} (x - 4)^{\frac{3}{2}} + 4 x - 4) + C

\frac{\partial}{\partial x} (r^{2})

\int e^{- s t} \cdot e^{a t} d t

f (x) = \frac{2}{3 x + 5}, (- 1, - 1)

f (x) = x^{2} - 2 x, at x = - 1

y^{^{'}} + 5 y = 2 sin (e^{5 x})

What is the integral of x/(sqrt(x-4)) ?
The integral of x/(sqrt(x-4)) is 2(1/3 (x-4)^{3/2}+4sqrt(x-4))+C